Question

[Dương Hải Minh]

CMR : \(\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{n^2}< 1\) , Với n ∈ N , n > 1

Answer 1

[Phùng Khánh Linh]

Mạn phép sửa lại đề : \(\dfrac{1}{2^2}+\dfrac{1}{3^3}+...+\dfrac{1}{n^2}\)

Ta nhận thấy : \(\dfrac{1}{k^2}=\dfrac{1}{k.k}< \dfrac{1}{k.\left(k-1\right)}\) Vì : k > k - 1

Lại có : \(\dfrac{1}{k\left(k-1\right)}=\dfrac{1}{k-1}-\dfrac{1}{k}\)

Ta có :

\(\dfrac{1}{2^2}+\dfrac{1}{3^3}+...+\dfrac{1}{n^2}\) < \(\dfrac{1}{2\left(2-1\right)}+\dfrac{1}{3\left(3-1\right)}+...+\dfrac{1}{n\left(n-1\right)}\)

\(\dfrac{1}{2^2}+\dfrac{1}{3^3}+...+\dfrac{1}{n^2}\) < \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\)

Do : n > 1 , nên : \(\dfrac{1}{2^2}+\dfrac{1}{3^3}+...+\dfrac{1}{n^2}\) < 1

Answer 2

[Kookie BTS]

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